1 Introduction motivation
The limit is well known in the literature as the Keller’s limit, see [2]. Such a limit is very useful in many mathematical contexts and contributes as a tool for establishing some interesting inequalities [3‐6].
$$\lim_{n\rightarrow\infty} \biggl(\frac{(n+1)^{n+1}}{n^{n}}-\frac{n^{n}}{(n-1)^{n-1}} \biggr)=e $$
In the recent paper [1], Mortici et al. have constructed a new proof of the limit and have discovered the following new results which generalize the Keller limit.
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Theorem 1
Let
c
be any real number and let
Then
$$u_{n}(c)=(n+1) \biggl(1+\frac{1}{n+c} \biggr)^{n+c}-n \biggl(1+\frac {1}{n+c-1} \biggr)^{n+c-1}-e. $$
$$\begin{aligned}& \lim_{n\rightarrow\infty}u_{n}(c)=0, \end{aligned}$$
(1.1)
$$\begin{aligned}& \lim_{n\rightarrow\infty}n^{2}u_{n}(c)= \frac{e}{24}(1-12c), \end{aligned}$$
(1.2)
$$\begin{aligned}& \lim_{n\rightarrow\infty}n^{3}u_{n}\biggl( \frac{1}{12}\biggr)=\frac{5e}{144}. \end{aligned}$$
(1.3)
The proof of Theorem 1 given in [1] is based on the following double inequality for every x in \(0< x\leq1\): where and
$$a(x)< (1+x)^{1/x}< b(x), $$
$$a(x)=e-\frac{e}{2x}+\frac{11ex^{2}}{24}-\frac{21ex^{3}}{48}+ \frac {2{,}447ex^{4}}{5{,}760}-\frac{959ex^{5}}{2{,}304} $$
$$b(x)=a(x)+\frac{959ex^{5}}{2{,}304}. $$
But, this proof has a major objection, namely, for the reader it is very difficult to observe the behavior of \(u_{n}(c)\) as \(n\rightarrow\infty\).
In this note, we will establish an integral expression of \(u_{n}(c)\), which tells us that Theorem 1 is a natural result.
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2 Main results
To establish an integral expression of \(u_{n}(c)\), we first recall the following result we obtained in [7].
Theorem 2
Let
\(h(s)=\frac{\sin(\pi s)}{\pi}s^{s}(1-s)^{1-s}\), \(0\leq s\leq 1\). Then for every
\(x>0\), we have
where
$$ \biggl(1+\frac{1}{x} \biggr)^{x} =e \Biggl(1-\sum _{j=1}^{\infty}\frac{b_{j}}{(1+x)^{j}} \Biggr), $$
(2.1)
$$\begin{aligned}& b_{1}=\frac{1}{2}, \end{aligned}$$
(2.2)
$$\begin{aligned}& b_{j}=\frac{1}{e} \int_{0}^{1}s^{j-2}h(s)\,ds \quad(j=2,3, \ldots). \end{aligned}$$
(2.3)
Hence
$$\begin{aligned}& \int_{0}^{1}h(s)\,ds=\frac{e}{24}, \end{aligned}$$
(2.4)
$$\begin{aligned}& \int_{0}^{1}sh(s)\,ds=\frac{e}{48}. \end{aligned}$$
(2.5)
Now, we establish an integral expression of \(u_{n}(c)\). Equation (2.1) implies the following results:
Hence by (2.2), (2.3), (2.6), and (2.7), we have
$$\begin{aligned}& \biggl(1+\frac{1}{n+c} \biggr)^{n+c} =e \Biggl(1-\sum _{j=1}^{\infty}\frac{b_{j}}{(1+n+c)^{j}} \Biggr), \end{aligned}$$
(2.6)
$$\begin{aligned}& \biggl(1+\frac{1}{n+c-1} \biggr)^{n+c-1} =e \Biggl(1-\sum _{j=1}^{\infty}\frac{b_{j}}{(n+c)^{j}} \Biggr). \end{aligned}$$
(2.7)
$$\begin{aligned} u_{n}(c) =&\frac{e}{2} \biggl(\frac{n}{n+c} - \frac{n+1}{1+n+c} \biggr)+ \int_{0}^{1}h(s)\sum_{j=2}^{\infty} \frac {ns^{j-2}}{(n+c)^{j}}\,ds \\ &{}- \int_{0}^{1}h(s)\sum_{j=2}^{\infty} \frac{(n+1)s^{j-2}}{(1+n+c)^{j}}\,ds. \end{aligned}$$
(2.8)
Note that
Therefore, from (2.8)-(2.11), we obtain the desired result: where
$$\begin{aligned}& \frac{e}{2}= \int_{0}^{1}12h(s)\,ds, \end{aligned}$$
(2.9)
$$\begin{aligned}& \sum_{j=2}^{\infty} \frac{ns^{j-2}}{(n+c)^{j}}=\frac{1}{(n+c)(n+c-s)}, \end{aligned}$$
(2.10)
$$\begin{aligned}& \sum_{j=2}^{\infty} \frac{(n+1)s^{j-2}}{(1+n+c)^{j}}=\frac{1}{(1+n+c)(1+n+c-s)}. \end{aligned}$$
(2.11)
$$\begin{aligned} u_{n}(c)= \int_{0}^{1}h(s)\frac{(1-12c)n^{2} +(1-12c+24cs-24c^{2})n+K}{(n+c)(1+n+c)(n+c-s)(1+n+c-s)}\,ds, \end{aligned}$$
(2.12)
$$K=s^{2}-(1+2c)s+c+c^{2}. $$
From (2.12), we get immediately
$$\begin{aligned}& \lim_{n\rightarrow\infty}u_{n}(c)=0, \\& \lim_{n\rightarrow\infty}n^{2}u_{n}(c)= \int_{0}^{1}(1-12c)h(s)\,ds \\& \hphantom{\lim_{n\rightarrow\infty}n^{2}u_{n}(c)}=\frac{e}{24}(1-12c), \\& \lim_{n\rightarrow\infty}n^{3}u_{n}\biggl( \frac{1}{12}\biggr)= \int_{0}^{1}\biggl(2s-\frac{1}{6} \biggr)h(s)\,ds \\& \hphantom{\lim_{n\rightarrow\infty}n^{3}u_{n}\biggl( \frac{1}{12}\biggr)}=\frac{e}{24}-\frac{1}{6}\times\frac{e}{24} \\& \hphantom{\lim_{n\rightarrow\infty}n^{3}u_{n}\biggl( \frac{1}{12}\biggr)}=\frac{5e}{144}. \end{aligned}$$
3 Conclusions
Acknowledgements
The work was supported by the National Natural Science Foundation of China, No. 11471103.
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Competing interests
The authors declare that there is no conflict of interests regarding the publication of this article.
Authors’ contributions
The authors completed the paper together. They also read and approved the final manuscript.